The demand for cars from a car hire firm may be modelled by a Poisson distribution with mean $4$ per day.
- Find the probability that in a randomly chosen day the demand is for $3$ cars.
- The firm has $5$ cars available for hire. Find the probability that demand exceeds the number of cars available.
Mean of Poisson distribution $= \lambda = 4$
- $P \left(\text{Demand is for 3 cars}\right) = P \left(X = 3\right)$
$\begin{aligned} P \left(X = 3\right) & = \dfrac{e^{- \lambda} \times \lambda^{3}}{3!} \\\\ & = \dfrac{e^{- 4} \times 4^3}{6} \\\\ & = \dfrac{0.01832 \times 64}{6} = 0.1954 \end{aligned}$
- Number of cars available for hire $= 5$
$P \left(\text{Demand exceeds number of available cars}\right) = P \left(X > 5\right)$
$\begin{aligned} P \left(X > 5\right) & = 1 - P \left(\text{Demand is for 5 cars or less}\right) \\\\ & = 1 - P \left(X \leq 5\right) \end{aligned}$
$\begin{aligned} P \left(X \leq 5\right) & = \sum \limits_{x = 0}^{5} \dfrac{e^{- \lambda} \times \lambda^{x}}{x!} \\\\ & = e^{- \lambda} \times \left[\dfrac{\lambda^{0}}{0!} + \dfrac{\lambda^{1}}{1!} + \dfrac{\lambda^{2}}{2!} + \dfrac{\lambda^{3}}{3!} + \dfrac{\lambda^{4}}{4!} + \dfrac{\lambda^{5}}{5!} \right] \\\\ & = e^{- 4} \times \left[1 + \dfrac{4}{1} + \dfrac{4^2}{2} + \dfrac{4^3}{6} + \dfrac{4^4}{24} + \dfrac{4^5}{120}\right] \\\\ & = 0.01832 \times 42.8667 = 0.7853 \end{aligned}$
$\therefore \;$ $P \left(\text{Demand exceeds number of available cars}\right) = 1 - 0.7853 = 0.2147$